Question

The block of mass $m$ Is in equilibrium relative to the smooth wedge of mass $M$ Which is pushed by a horizontal force $F .$ Find

(a) Pseudo force acting on m

(b) $M$ As viewed by the observer sitting on the wedge. Will these pseudo forces

(c) equal and opposite, action–reaction pairs? Explain.

Accepted Answer

Taking block and wedge as system.

Net force on the system horizontally $= F$ .

Total mass $= M + m$ .

Therefore, acceleration of the system in $+ ve x$ -direction,

$a = \frac{F}{M + m} \Rightarrow \vec{a} = \frac{F}{M + m} \hat{i}$

(a) Pseudo force acting on block is in opposite direction of observer magnitude and acceleration equal to multiplication of mass of the object and the acceleration of the observer,

${\vec{F}}_{PS} = - m a \hat{i} = - \frac{m F}{M + m} \hat{i}$

(b) Pseudo force acting on block,

$M {\vec{F}}_{PS}^{'} = - M a \hat{i}$

${\vec{F}}_{PS}^{'} = - \frac{M F}{M + m} \hat{i}$

(c) Pseudo force is not a real force. It is applied to solve the problems of non-inertial frame. It does not have action–reaction pairs.

B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 41: ILLUSTRATION 6.37

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 41: ILLUSTRATION 6.37

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